Question

How do you solve the following system of equations $-2x+y=1,2x+2y=11$ ?

Answer 1

Hint: We are going to solve the given system of equations $-2x+y=1,2x+2y=11$ by using elimination method. In this method, we will first of all eliminate x by adding both the equations. After adding the two equations, we will be left with the equation in y so from that we will get the value of y. And then we will substitute this value of y in any of the two equations to get the value of x.

Complete step by step answer:
The system of equations which we have to solve is as follows:
$-2x+y=1,2x+2y=11$
Now, let us name the first equation as (1) and second as (2).
$\begin{align}
  & -2x+y=1.......(1) \\
 & 2x+2y=11......(2) \\
\end{align}$
Now, we are going to eliminate x by adding (1) and (2).
$\begin{align}
  & -2x+y=1 \\
 & \dfrac{+2x+2y=11}{0+3y=12} \\
\end{align}$
From the above addition, we have got the equation in y as follows:
$3y=12$
Dividing 3 on both the sides we get,
$\begin{align}
  & \dfrac{3y}{3}=\dfrac{12}{3} \\
 & \Rightarrow y=4 \\
\end{align}$
In the above, 3 will be cancelled out from the numerator and the denominator in the L.H.S and 12 will be divided 4 times by 3 on the R.H.S.
From the above addition, we got the value of y as 4. Now, we are substituting this value of y in (1) we get,
$-2x+4=1$
Subtracting 4 on both the sides we get,
$\begin{align}
  & -2x+4-4=1-4 \\
 & \Rightarrow -2x+0=-3 \\
\end{align}$
In the above equation, negative sign will be cancelled out from both the sides and we are left with:
$2x=3$
Now, dividing 2 on both the sides we get,
$\begin{align}
  & \dfrac{2x}{2}=\dfrac{3}{2} \\
 & \Rightarrow x=\dfrac{3}{2} \\
\end{align}$

From the above solution, we got the solutions of system of equations as:
$x=\dfrac{3}{2},y=4$

Note: In the above solution, you can check the solutions of the system of equations that we are getting are correct or not by substituting these values of x and y in any of the two equations.
The values of x and y that we are getting is equal to:
$x=\dfrac{3}{2},y=4$
Substituting the above values of x and y in (1) we get,
$-2\left( \dfrac{3}{2} \right)+\left( 4 \right)=1$
In the above equation, 2 will be cancelled out from the numerator and denominator in the L.H.S and we get,
$\begin{align}
  & -3+4=1 \\
 & \Rightarrow 1=1 \\
\end{align}$
As you can see that L.H.S is equal to R.H.S so the values of x and y that we have found above is correct.